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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Solid \(Ba(NO_3)_2\) is gradually dissolved in a \(1.0 \times 10^{-4}\, M\, Na_2CO_3\) solution. At which concentration of \(Ba^{2+}\), precipitate of \(BaCO_3\) begins to form ? \((K_{sp}\) for \(BaCO_3 = 5.1 \times 10^{-9})\)
- A \(5.1 \times {10^{ - 5}}\,M\)
- B \(7.1 \times {10^{ - 8}}\,M\)
- C \(4.1 \times {10^{ - 5}}\,M\)
- D \(8.1 \times {10^{ - 7}}\,M\)
Answer & Solution
Correct Answer
(A) \(5.1 \times {10^{ - 5}}\,M\)
Step-by-step Solution
Detailed explanation
Given \(N{a_2}C{O_3} = 1.0 \times {10^{ - 4}}\,M\) \(\therefore \,[CO_3^ - ] = 1.0 \times {10^{ - 4}}\,M\) \(i.e.\,\,\,s = 1.0 \times {10^{ - 4}}\,M\) At equilibrium \([B{a^{ + + }}][CO_3^ - ] = {K_{sp}}\,of\,BaC{O_3}\)…
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